Question 1122568: Prove if 0 < a ≤ b, then a^2 ≤ b^2 and √a ≤√b.
Thanks in advance for the help.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! There are two cases to consider
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(1) if a = b, then a^2 = b^2 and sqrt(a) = sqrt(b)
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(2) if a < b, then b-a > 0 and b+a > 0. By closure under addition
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(b-a)(b+a) > 0 and b^2 > a^2 or a^2 < b^2
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a > 0 then b > 0, a < b then take sqrt of both sides of "<", sqrt(a) < sqrt(b)
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